Problem: Let $g(x)=\begin{cases} \dfrac{x^2+5x+6}{x+3}&\text{for }x\neq -3 \\\\ k&\text{for }x=-3 \end{cases}$ $g$ is continuous for all real numbers. What is the value of $k$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-6$ (Choice B) B $-2$ (Choice C) C $-3$ (Choice D) D $-1$
Solution: $\dfrac{x^2+5x+6}{x+3}$ is continuous for all real numbers other than $x=-3$ which means $g$ is continuous for all real numbers other than $x=-3$. In order for $g$ to also be continuous at $x=-3$, the following equality must hold: $\lim_{x\to -3}g(x)=g(-3)$ Since $g(-3)=k$, we will obtain the above equality by letting $k=\lim_{x\to -3}g(x)$. So let's find $\lim_{x\to -3}g(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to -3}g(x) \\\\ &=\lim_{x\to -3}\dfrac{x^2+5x+6}{x+3} \gray{\text{This is the rule for }x\neq -3} \\\\ &=\lim_{x\to -3}\dfrac{\cancel{(x+3)}(x+2)}{\cancel{(x+3)}} \gray{\text{Factor}} \\\\ &=\lim_{x\to -3}{(x+2)} \gray{\text{Cancel common factors}} \\\\ &\text{(This is allowed because }x\neq -3) \\\\ &=(-3)+2 \gray{\text{Direct substitution}} \\\\ &=-1 \end{aligned}$ We obtained that if we set $k=-1$, then $\lim_{x\to -3}g(x)=g(-3)$, which makes $g$ continuous at $x=-3$. Since we already saw that $g$ is continuous for any other real number, we can determine that it's continuous for all real numbers. In conclusion, $k=-1$.